find the directional derivative of f(x y z)=x^2+y^2+z^3 at the point (1,2,3) in the direction of the line x/2+y/2+z/2
Answers
Answer:
Step-by-step explanation:
To find a directional derivative we note that Duf = ∇f · u. So we need to
find u and ∇f. Since we are going from the point (2, −6, 3) towards (0, 0, 0)
then a vector pointing in the appropriate direction is h−2, 6, −3i. This is not
a unit vector since kh−2, 6, −3ik =
√
4 + 36 + 9 = √
49 = 7, but by scaling
we can make it a unit vector so that u =
1
7
h−2, 6, −3i. The gradient is
∇f(x, y, z) = hz
2 − 3y + 2yz − 3, −3x + 2xz + 5, 2xz + 2xyi.
Evaluating at the point (2, −6, 3) we have
∇f(2, −6, 3)
= h3
2 − 3·(−6) + 2·(−6)·3 − 3, −3·2 + 2·2·3 + 5, 2·2·3 + 2·2·(−6)i
= h−12, 11, −12i.
Therefore the desired directional derivative will be given by
∇f · u = h−12, 11, −12i · 1
7
h−2, 6, −3i =
24 + 66 + 36
7
=
126
7
= 18.